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Living_dreams

  • 3 years ago

find all local max,min and/or saddle point of f(x,y)=4+x^3+y^-3xy

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  1. Avva
    • 3 years ago
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    would you plz write the function using equation

  2. amistre64
    • 3 years ago
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    hmm, something to do with:\[f_{xy}f_{yx}-(f_{xx})^2\] or did i recall that incorrectly?

  3. amistre64
    • 3 years ago
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    almost had it :) \[f_{xx}f_{yy}-(f_{xy})^2\]

  4. Living_dreams
    • 3 years ago
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    \[4+x ^{3}+y ^{3}-3xy\]

  5. Living_dreams
    • 3 years ago
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    it fxx.fyy-(fxy)^2

  6. Living_dreams
    • 3 years ago
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    i jst want to verify the answer for it !

  7. TuringTest
    • 3 years ago
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    what answer did you get?

  8. Avva
    • 3 years ago
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    \[fx = 2x-3y \] \[fy= 2y-3x\] on solving both equations the point is (0,0) \[fxx = 2\] , \[fyy= 2\] \[fxy = \] \[D = -5 <0 saddle point\]

  9. TuringTest
    • 3 years ago
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    that's not quite right I don't think

  10. Avva
    • 3 years ago
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    fxy = -3

  11. TuringTest
    • 3 years ago
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    \[f(x,y)=4+x^3+y^3-3xy\]\[f_x=3x^2-3y=0\implies y=x^2\]\[f_y=3y^2-3x=3x^4-3x=3x^3(x-1)=0\]\[ x=\{0,1\},y=\{0,1\}\]

  12. Living_dreams
    • 3 years ago
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    its x^3 so i have (0,0) ,(1,1)and a (-1,1) i am not sure if (-1,1) should be there oor not

  13. Avva
    • 3 years ago
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    @Living_dreams I solved what you mentioned in your comment

  14. TuringTest
    • 3 years ago
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    there is no x=-1 solution, I don't see how you get that

  15. TuringTest
    • 3 years ago
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    @Avva I don't see how you get your fx and fy

  16. Avva
    • 3 years ago
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    Ooh I saw the powers on X ^ y 2 not 3

  17. Living_dreams
    • 3 years ago
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    @Avva thanks for ur help but i was wonder abt the max and min points! n\ @TuringTest : 3y^2-3x=0 i sub y=x^2 from other equation so it will be 3x(x^3-1)=0 therefore you have x =0,1,-1 i am not sure abt -1 though !! tht was my question

  18. TuringTest
    • 3 years ago
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    x^3=1 does not have x=-1 as a solution since (-1)^3=-1, not 1

  19. TuringTest
    • 3 years ago
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    I did screw up my factoring earlier though, you were right about that :P

  20. TuringTest
    • 3 years ago
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    3x(x^3-1)=0 3x=0 -> x=0 x^3=1 -> x=1

  21. Living_dreams
    • 3 years ago
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    @TuringTest that alrite!! :) so now my question is x^3=1 so will i hv x=-1 if i solve for x ?

  22. TuringTest
    • 3 years ago
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    no, again, because \((-1)^3\neq1\) so \(x^3=1\) has only one real solution, x=1 x=-1 is NOT a solution

  23. Avva
    • 3 years ago
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    the other two roots are complex not real

  24. Living_dreams
    • 3 years ago
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    @TuringTest oh ok thanks a lot for ur help!! so the final ans will be (0,0 ) is a saddle point and (1,1) will be a local minimum !

  25. TuringTest
    • 3 years ago
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    exactly :)

  26. Living_dreams
    • 3 years ago
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    @Avva :thanks for ur help !! :)

  27. Avva
    • 3 years ago
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    @Living_dreams URW :) sorry I didn't get that the powers are 3 at first

  28. Living_dreams
    • 3 years ago
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    it okay !!i knw it get hard to tell the difference between 2 and 3 if we type it in equation form !!

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